A man in a balloon, rising vertically with an acceleration of `5 m//s^(2)`, releases a ball 10 s after the balloon is let go from the ground. The greatest height above the ground reached by the ball is

Options

A. 125 m
B. 250 m
C. 375 m
D. 500 m

Explanation

To solve the problem step by step, we need to determine the maximum height reached by the ball after it is released from the balloon. Here’s how we can approach this:

### Step 1: Determine the velocity of the balloon at the time the ball is released
The balloon rises with an acceleration of \(5 \, \text{m/s}^2\). The time before the ball is released is \(10 \, \text{s}\).

Using the equation of motion:
\[ v = u + at \]
where:
- \(u = 0 \, \text{m/s}\) (initial velocity of the balloon)
- \(a = 5 \, \text{m/s}^2\) (acceleration of the balloon)
- \(t = 10 \, \text{s}\)

Substituting the values:
\[ v = 0 + (5 \times 10) = 50 \, \text{m/s} \]

### Step 2: Calculate the height of the balloon when the ball is released
We can use the equation of motion to find the height \(h\) of the balloon at the time of release:
\[ h = ut + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ h = 0 \times 10 + \frac{1}{2} \times 5 \times (10)^2 = \frac{1}{2} \times 5 \times 100 = 250 \, \text{m} \]

### Step 3: Determine the maximum height reached by the ball after it is released
After the ball is released, it has an initial upward velocity of \(50 \, \text{m/s}\) and is acted upon by gravity, which decelerates it at \(g = 10 \, \text{m/s}^2\).

Using the equation:
\[ v^2 = u^2 + 2a s \]
where:
- \(v = 0 \, \text{m/s}\) (final velocity at the maximum height)
- \(u = 50 \, \text{m/s}\) (initial velocity of the ball)
- \(a = -10 \, \text{m/s}^2\) (acceleration due to gravity, negative since it acts downward)
- \(s\) is the height gained after release.

Rearranging gives:
\[ 0 = (50)^2 + 2(-10)s \]
\[ 0 = 2500 - 20s \]
\[ 20s = 2500 \]
\[ s = \frac{2500}{20} = 125 \, \text{m} \]

### Step 4: Calculate the total maximum height above the ground
The total height \(H\) above the ground when the ball reaches its maximum height is the sum of the height of the balloon when the ball was released and the height gained by the ball after release:
\[ H = h + s = 250 \, \text{m} + 125 \, \text{m} = 375 \, \text{m} \]

### Final Answer
The greatest height above the ground reached by the ball is **375 meters**.

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